1.4k views

Consider the following C program:

void abc(char*s)
{
if(s[0]=='\0')return;
abc(s+1);
abc(s+1);
printf("%c",s[0]);
}

main()
{
abc("123");
}
1. What will be the output of the program?
2. If $abc(s)$ is called with a null-terminated string $s$ of length $n$ characters (not counting the null ('\0') character), how many characters will be printed by $abc(s)$?
edited | 1.4k views

Answer is a) $332 \ 332 \ 1$
Answer b) $2^n-1$

Q. a) $O/p$ :

3 3 2 3 3 2 1

b) : $T(n) = 2T(n-1) +1 ; n>0$

$= 0 ; n=0$ [Since for length zero string no character will be printed]

After solving it by substitution,

$T(n) = 2T(n-1) +1$

$= 2(2T(n-2) + 1 ) +1$

$= 2^2T(n-2) + 2 +1$

$= 2^2(2T(n-3) +1 ) + 2 + 1$

$= 2^3T(n-3) + 2^2 + 2 + 1$

Finally, it will expand like

$= 2^nT(n-n) + 2^{n-1} + 2^{n-2} + \ldots + 2^2 + 2 + 1$

$= 2^nT(0) + 2^{n-1} + 2^{n-2} + \ldots+ 2^2 + 2 + 1$

$= \frac{1.(2^n -1)}{(2 - 1)}$

$= 2^n -1$

edited
0
can you plz explain more clearly ... as i am not getting the meaning of the code line ...??
0
How did u get the recurrence relation ? I am confused .It should be T(n) = 2* T(n+1) +1. Plz help.
+1
we are decreasing length of input by 1 so that's why n-1 in recurrence relation.
0
how it can possible that we have problm of size n and after diving it  into subprobem it size gets incresed . its not possible , but if u r willing to increase the size of problem in each call then problem will never end.
0
its not possible one other way of checking is that solve ur recurrence relation it is giving time complexity in negative which is contradiction
ans for B is: 2^n -1

for length 2 -> 3 characters

for length 3-> 7 characters

for length 4-> 15 characters

hence for n-> 2^n - 1 characters
+4

a) 3323321

b) we can also get the answer of b by solving recurrence relation

T(n) = 2 * T(n-1) +1 ; n>0

=  0 ; n=0 [Since for length zero string no character will be printed]

After solving it by substitution,

T(n) = 2 * T(n-1) +1

= 2  x  (2  x  T(n-2)  + 1 )  +1

= 2x T(n-2)  +  2 +1

=  22 x (2*T(n-3) +1 )  + 2 + 1

= 2x T(n-3) + 22  + 2 + 1

Finally it will expand like

= 2x T(n-n) + 2n-1 + 2n-2 + - - - -  - - - + 2+ 2 + 1

= 2x T(0) + 2n-1 + 2n-2 + - - - - - - - - - - + 2+ 2 + 1

= 1x (2-1) / (2 - 1 )

= 2-1

0
But how do we come to this recurrence  relation first....i mean it works for all questions like this....do we have to learn this...please reply?
+3

lets assume to solve length n input it takes T(n) times

now in ques they have two abc(s+1) and one printf

abc(s+1) reduces the length of input by 1 so now it have length (n-1)

now to solve length (n-1) input it takes T(n-1) times and printf take 1 unit of time

hence reccurence relation is :

T(n)=2.T(n-1)+1

1
2